AP Calculus BC 2.1 Defining Average and Instantaneous Rates of Change at a Point - MCQs - Exam Style Questions
No-Calc Question
The function \(f\) is differentiable and has values as shown.
| \(x\) | 0 | 2 | 6 | 12 | 20 |
| \(f(x)\) | \(-8\) | \(-2\) | 0 | 2 | 18 |
On which interval is the average rate of change of \(f\) greatest?
(A) \([0,2]\)
(B) \([2,6]\)
(C) \([6,12]\)
(D) \([12,20]\)
(B) \([2,6]\)
(C) \([6,12]\)
(D) \([12,20]\)
▶️ Answer/Explanation
Detailed solution
Average rate \(\displaystyle \frac{f(b)-f(a)}{b-a}\).
\([0,2]: \frac{-2-(-8)}{2-0}=\frac{6}{2}=3\).
\([2,6]: \frac{0-(-2)}{4}=\frac{2}{4}=\tfrac12\).
\([6,12]: \frac{2-0}{6}=\tfrac13\).
\([12,20]: \frac{18-2}{8}=2\).
The greatest is \(3\) on \([0,2]\).
✅ Answer: (A)
▶️ Answer/Explanation
Solution
Given \( f(x) = (x^2 + 1)^3 \), compute \( \lim_{x \to -1} \frac{f(x) – f(x+1)}{x+1} \).
Evaluate at \( x = -1 \): \( f(-1) = (1 + 1)^3 = 8 \), \( f(x+1) = f(0) = (0 + 1)^3 = 1 \), so \( \frac{8 – 1}{-1 + 1} = \frac{7}{0} \), indicating a limit form.
Rewrite: \( \frac{f(x) – f(-1)}{x – (-1)} = \frac{(x^2 + 1)^3 – 8}{x + 1} \).
Use difference of cubes: \( (x^2 + 1)^3 – 8 = ((x^2 + 1) – 2)((x^2 + 1)^2 + 2(x^2 + 1) + 4) \), where \( x^2 + 1 – 2 = (x – 1)(x + 1) \).
Simplify: \( (x – 1)((x^2 + 1)^2 + 2(x^2 + 1) + 4) \).
At \( x = -1 \): \( (-1 – 1)(((-1)^2 + 1)^2 + 2((-1)^2 + 1) + 4) = (-2)(4 + 4 + 4) = -24 \).
✅ Answer: A)